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For $y\prime \prime +py\prime +qy=0$, the Wronskian satisfies $W\prime =-pW$.

More precisely, let $y_1,y_2$ be solutions of a differential equation for the same homogeneous linear differential equationon an interval $I$ where $p,q$ are continuous functions. For any $x_0\in I$, Abel's identity givesOnly the coefficient of $y^{\prime }$ enters the evolution of the Wronskian. In an unnormalized equation $A(x)y^{\prime \prime }+B(x)y^{\prime }+C(x)y=0$, first use $p=B/A$ on an interval where $A\ne 0$ and the coefficient ratios are continuous functions.^[ref]

The proof uses cancellation. Differentiating the Wronskian by the product rule and substituting the common equation givesMultiplying this first-order linear differential equation by the integrating factor $\mu (x)=\exp ({\int }_{x_0}^{x}p(s)ds)$ gives $(\mu W{)}^{\prime }=0$. Since $\mu (x_0)=1$, the constant is $W(x_0)$, proving Abel's identity. This argument also covers $W(x_0)=0$ without dividing by $W$.

The exponential function is positive for real $p$, so a real Wronskian either vanishes everywhere or has a fixed nonzero sign throughout $I$. In the preceding example, $p=-1$ and $W(0)=3$ give $W(x)=3e^x$ immediately. If $p=0$, every Wronskian of two solutions of a differential equation for that same equation is constant, regardless of $q$. Thus Abel's identity can certify linear independence and find the denominator used in variation of parameters without explicit formulas for both solutions of a differential equation.

The underlying graph of a finite Dynkin diagram is a forest. Normalize every simple root to length one. Each bonded pair has inner product at most $-1/2$, and every other distinct pair has nonpositive inner product. A cycle with $m$ vertices would make the squared norm of their sum at most $m-m=0$, contradicting their linear independence in an inner product space. Multiple bonds denote root-length information, not graph-theoretic parallel-edge cycles.

Points $p_0,\dots ,p_k$ in an affine space are affinely independent if the only affine dependence displayed above has all coefficients zero, using any choice of origin to read the vector equation. Since the coefficients sum to zero, replacing the origin leaves that equation unchanged. Substituting ${\lambda }_0=-{\sum }_{i=1}^k{\lambda }_i$ shows that this condition is equivalent to linear independence of $p_1-p_0,\dots ,p_k-p_0$. It also says that an affine combination of the points has unique coefficients. In ${\mathbb{R}}^n$, at most $n+1$ points can be affinely independent; $n+1$ such points span a full-dimensional simplex.

Every linearly independent subset of a finite-dimensional vector space extends to a basis. Starting with its vectors, add a vector outside their current span until the span is the whole space. Each addition preserves linear independence, and the process stops because the dimension is finite. This also lets a linear functional on a vector subspace extend to the ambient space by assigning arbitrary values on the additional basis vectors.

A Chebyshev system on a nondegenerate real interval is a family of $n+1$ real continuous functions such that every linear combination with coefficients not all zero has at most $n$ distinct zeros. The condition entails linear independence and unique interpolation at any $n+1$ distinct points. It is a zero-counting hypothesis, not a requirement to count multiplicities.

A scalar multiplying a specified term in a linear combination, polynomial or power series. When the term family has linear independence, the coefficients are unique; uniqueness of finite Laurent polynomial coefficients follows after multiplying by a sufficiently large power to obtain an ordinary polynomial.

Let $A,B$ be unital associative algebras over a field, each a simple ring, and $J\ne 0$ a two-sided ideal of their tensor product. Choose $0\ne x={\sum }_{i=1}^n{a}_i\otimes b_i\in J$ of minimal two-factor tensor rank. Both factor families are linearly independent. Since $A{a}_{1}A=A$, choose ${\sum }_j{u}_j{a}_1{v}_j=1$ and replace $x$ by ${\sum }_j(u_j\otimes 1)x(v_j\otimes 1)$. The result has first coefficient $1$, is nonzero by the linear independence of the $b_i$, and retains minimal two-factor tensor rank; its first factor family is therefore still linearly independent. Now apply the same construction in $B$ to make the first second-factor coefficient $1$. The result is again nonzero of minimal two-factor tensor rank, and both its coefficient families are linearly independent.

If $Z(A)=k$, commutation with $a\otimes 1$ gives a tensor in $J$ expressible with at most $n-1$ pure tensors, hence zero by minimality. Independence forces every $c_i$ to be central, hence a scalar multiple of $1$. Independence of $1,c_2,\dots ,c_n$ then forces $n=1$. Thus $1\otimes 1\in J$, proving simplicity of the tensor product without a splitting-field argument.

A constraint qualification is a regularity hypothesis on optimization constraints that guarantees a specified optimality or duality conclusion. For the necessity of Karush-Kuhn-Tucker conditions, an appropriate constraint qualification ensures that a local minimum admits suitable Lagrange multipliers; linear independence of the gradients of equality optimization constraints and active constraints is one sufficient hypothesis in KKT regularity at a local minimum.^[ref] Slater's condition provides a different constraint qualification for convex optimization, yielding strong duality and dual attainment.^[ref]

The unique scalar coefficients in the finite linear combination expressing a vector in a chosen basis. Spanning gives existence, while linear independence makes the difference of two proposed representations have all zero coefficients. Ordering a finite basis turns these coefficients into a coordinate vector.

On a $d$-dimensional vector space, a cyclic vector $v$ gives a basis $v,Tv,\dots ,T^{d-1}v$. Write $T^{d}v=-{\sum }_{j<d}{c}_j{T}^{j}v$ and set $f(t)=t^d+{\sum }_{j<d}{c}_j{t}^j$. Then $f(T)v=0$. Since polynomials in $T$ commute with $T$, $f(T)$ kills every basis vector, so $f(T)=0$. No nonzero polynomial of degree less than $d$ can kill $v$, by linear independence. Hence the minimal polynomial is $f$. The matrix in this basis is $C(f)$, and the characteristic polynomial of a companion matrix is $f$ as well.

Let $T:U\to V$ be a linear map between finite-dimensional vector spaces and $X$ a vector subspace of $V$. The preimage $T^{-1}(X)$ is a vector subspace containing $\ker T$. To compute its dimension, choose a basis $k_1,\dots ,k_s$ of $\ker T$ and a basis $y_1,\dots ,y_d$ of $X\cap \mathrm{im}T$. Lift each $y_j$ to $u_j\in U$ with $T{u}_j=y_j$. Then $k_1,\dots ,k_s,u_1,\dots ,u_d$ is a basis of $T^{-1}(X)$: applying $T$ proves linear independence, and subtracting a suitable linear combination of the lifts reduces every element of the preimage to the kernel. This proves the displayed formula without using the rank-nullity theorem. Taking $X=V$ then proves the rank-nullity theorem itself.

For bases $(u_i)$ of $U$ and $(v_j)$ of $W$, the displayed pairs form a basis of $U\oplus W$. Expanding each component proves spanning; comparing components of a vanishing linear combination proves linear independence. Therefore $\dim (U\oplus W)=\dim U+\dim W$ for two finite-dimensional vector spaces.

For two solutions of the linear recurrence relation $x_{n+2}+b_n{x}_{n+1}+c_n{x}_n=0$, their discrete Wronskian obeys $W_{n+1}=c_n{W}_n$. Substitute the recurrence into $p_{n+1}{q}_{n+2}-p_{n+2}{q}_{n+1}$: the $b_n$ terms cancel and the remaining expression is $c_n(p_n{q}_{n+1}-p_{n+1}{q}_n)$. Iteration gives $W_{n+1}=W_1{\prod }_{m=1}^n{c}_m$, including when some $c_m$ vanish. A nonzero discrete Wronskian proves linear independence of the two solutions.

Let $V$ have finite dimension $n$ and let $S\subseteq V$ have $n$ vectors. Then spanning, linear independence and being a basis are equivalent. From a spanning set, remove redundant vectors until an independent spanning set remains. A removal would give a basis of fewer than $n$ vectors, contradicting dimension invariance. Conversely, if $S$ is independent and a vector lies outside its span, adjoining it gives $n+1$ independent vectors, again impossible. Thus an independent $S$ spans.

The $k$th exterior power ${\Lambda }^{k}V$ is the quotient vector space of the tensor power $V^{\otimes k}$ that makes every elementary tensor with two equal factors zero. More precisely, take the quotient vector space by the linear span of all such tensors. Its elementary products $v_1\wedge \cdots \wedge v_k$ turn alternating multilinear maps into linear maps: every alternating $k$-linear map on $V$ factors uniquely through ${\Lambda }^{k}V$.^[ref]

If $e_1,\dots ,e_n$ is an ordered basis, the increasing productsform a basis of ${\Lambda }^{k}V$. There is one for each $k$-element choice of basis directions, soRepeated directions vanish; products with a different order can be sorted using their sign. Coordinate determinants distinguish the remaining increasing products, establishing linear independence.^[ref]

For $V={\mathbb{R}}^3$, a basis of ${\Lambda }^{2}V$ is $e_1\wedge e_2,e_1\wedge e_3,e_2\wedge e_3$. This explains the three bivector components in the worked example. The top exterior power ${\Lambda }^{3}V$ has one basis element, $e_1\wedge e_2\wedge e_3$, while ${\Lambda }^{4}V=0$: four independent directions cannot fit into that vector space. By convention ${\Lambda }^{0}V$ is the scalar field, even when $V$ has dimension zero.

For a real $n$-dimensional vector space of continuous functions on a compact interval, choose sites maximizing the absolute evaluation determinant in any basis. Linear independence ensures some nonzero evaluation determinant, so a maximizer exists and has distinct sites. Replacing row $i$ by evaluation at $t$ multiplies the determinant by the cardinal function ${\ell }_i(t)$. Maximality implies $|{\ell }_i(t)|\le 1$, hence the Lebesgue constant of interpolation is at most $n$. This gives a useful interpolation set without asserting that determinant maximization minimizes the Lebesgue constant of interpolation.

For an explicit basis extension in ${\mathbb{R}}^2$, start with $A=\{u\}$, where $u=(1,1)$. Its linear span is the line $y=x$. Adding $w=(1,-1)$ preserves linear independence, because $w$ is outside that line, and gives a basis. Every vector has the finite linear combinationThe vector $2u=(2,2)$ cannot be added while preserving linear independence: it is already in the linear span. Many choices outside the line work, so maximality does not select a unique basis.

For a finite-dimensional vector space with a supplied finite spanning set, a greedy algorithm scans the list, retaining each vector precisely when it is outside the linear span of those already retained. Starting with a linearly independent $A$, the same scan adjoins only the needed vectors. Retained vectors remain linearly independent, and every skipped vector is in the final linear span, so the result is a basis containing $A$. Finite reasoning needs no axiom of choice.

For an arbitrary vector space, this finite stopping argument is unavailable. Accepting Zorn's lemma asserts existence of a maximal element once every chain in a partial order has an upper bound; it supplies no general computational recipe. Over ZF, Zorn's lemma, the axiom of choice and the well-ordering theorem are equivalent. Particular infinite vector spaces can still have explicit bases, such as $1,x,x^2,\dots$ for the vector space of univariate polynomials over a specified field; infinity alone does not force a nonconstructive argument.^[ref]

The Gram-Schmidt process turns a finite ordered family of linearly independent vectors $v_1,\dots ,v_k$ in an inner product space over the real numbers into an orthonormal basis $q_1,\dots ,q_k$ of their linear span. It keeps every successive linear span, so the first $j$ new vectors describe exactly the same directions as the first $j$ inputs. Its idea is geometric: remove the orthogonal projection onto directions already found, then divide the remaining vector by its norm. This makes a complicated basis easier to use without changing the vector subspace it describes.^[ref]

Start with $u_1=v_1$ and $q_1=v_1/\Vert v_1\Vert$. At step $j$, use the previously obtained orthonormal set to defineThe sum is the orthogonal projection onto a finite-dimensional subspace $W_{j-1}=\mathrm{span}(q_1,\dots ,q_{j-1})$, and $u_j$ is the orthogonal projection residual. Linear independence ensures $u_j\ne 0$: otherwise $v_j$ would already be a linear combination of earlier inputs. If the input family is linearly dependent, use only the retained orthonormal vectors in the sum and omit an input whenever its remainder is the zero vector; do not divide by zero. The retained outputs still form an orthonormal basis of the original linear span, but their number can be smaller than $k$.

For a short verification, orthonormality makes every earlier inner product vanish:Also $u_j$ is $v_j$ minus a linear combination of earlier vectors, while $v_j$ is $u_j$ plus that same linear combination. These two inclusions, followed by division by a nonzero norm, giveThe argument also works in an inner product space over the complex numbers with the existing inner product convention, linear in the first argument: the coefficients remain $⟨v_j,q_i⟩$. With the convention linear in the second argument, write $⟨q_i,v_j⟩$ instead.

For a first picture, take $v_1=(1,1{)}^T$ and $v_2=(1,0{)}^T$ in ${\mathbb{R}}^2$ with the ordinary dot product. ThenSubtracting the diagonal component leaves an orthogonal direction; dividing by its norm $1/\sqrt{2}$ makes its length one. If the second input were $(2,2{)}^T$ instead, its orthogonal projection would be the entire input and $u_2=0$, revealing linear dependence.

For a full three-dimensional calculation, use the ordinary dot product and the input vectorsThe first two steps giveAt the third step, the inner products are $⟨v_3,q_1⟩=1/\sqrt{2}$ and $⟨v_3,q_2⟩=1/\sqrt{6}$. ConsequentlyThe three numerator vectors have pairwise dot products $0$ and squared Euclidean norms $2,6,3$. Thus the displayed outputs are indeed an orthonormal basis, and the three nonzero remainders also confirm linear independence of the inputs. Keeping the removed coefficients produces the QR decomposition below.

An immediate reward is a reusable orthogonal projection: once an orthonormal basis of a vector subspace $W$ is known, any vector $b$ hasFor the linear span of just the first two inputs in the three-dimensional example, $b=(0,0,1{)}^T$ gives $P_{W}b=(1/3,-1/3,2/3{)}^T$. Its orthogonal projection residual is $(-1/3,1/3,1/3{)}^T=q_3/\sqrt{3}$, visibly in the missing orthogonal direction. This is the geometry behind the linear least-squares problem.

The formulas above describe exact arithmetic. With floating-point arithmetic, subtracting an orthogonal projection nearly equal to the input can leave a small remainder; dividing by its small norm can magnify existing roundoff error. The Classical Gram-Schmidt and Modified Gram-Schmidt arrangements agree in exact arithmetic but can give different rounded results. Exact linear dependence and a small computed remainder therefore require different judgments in a numerical implementation.^[ref]

A matroid independent set is a subset belonging to the family $\mathcal{I}$. Independence refers to the chosen matroid: for a linear matroid it means linear independence, and for a graphic matroid it means that the chosen edges contain no graph cycle. It is different from an independent set, which selects vertices with no edge between them.

Let $m,n$ be coprime square-free integers, $m\equiv 3(\mathrm{mod}4)$ and $n\equiv 2(\mathrm{mod}4)$. For the biquadratic extension $L=\mathbb{Q}(\alpha ,\beta )$, with ${\alpha }^2=m$ and ${\beta }^2=n$, an integral basis is $1,\alpha ,\beta ,(\beta +\alpha \beta )/2$. To prove spanning, write an algebraic integer as $(a+b\alpha +c\beta +d\alpha \beta )/2$. The field traces to the three quadratic subfields and the ring of integers of a quadratic field formula force $a,b,c,d\in \mathbb{Z}$. The field norm to $\mathbb{Q}(\alpha )$ has coefficients $(a^2+m{b}^2-n{c}^2-mn{d}^2)/4$ and $(ab-ncd)/2$. Their integrality gives $ab\equiv 0(\mathrm{mod}2)$ and $a^2-b^2-2(c^2+d^2)\equiv 0(\mathrm{mod}4)$. Thus $a,b$ are even and $c\equiv d(\mathrm{mod}2)$, exactly the asserted integer span. Conversely, the half-sum $w$ has $w^2=n(1+m)/4+(n/2)\alpha \in \mathbb{Z}[\alpha ]$, so integral dependence is transitive proves it is an algebraic integer. Rational linear independence is immediate from its nonzero $\alpha \beta$ coefficient.